FP1 Mark Schemes from old P4, P5, P6 and FP1, FP2, FP3 papers (back to June 2002)

Please note that the following pages contain mark schemes for questions from past papers which were not written at an AS standard and may be less accessible than those you will find on future AS FP1 papers from Edexcel. Some questions would certainly worth more marks at AS level.

The standard of the mark schemes is variable, depending on what we still have – many are scanned, some are handwritten and some are typed.

The questions are available on a separate document, originally sent with this one.

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 1

[P4 January 2002 Qn 1]

[P4 January 2002 Qn 3]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 2

2.

[P4 January 2002 Qn 4]

5. 6r2 – 6 = n(n + 1)(2n + 1) , – 6n M1, A1

= n(2n2 + 3n – 5) M1

= n(n – 1)(2n – 5) (*) A1

(4 marks)[P4 June 2002 Qn 1]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 3

4.

[*P4 January 2002 Qn 5]

3.

6. (a) |w| = 50 (or equivalent) B1 (1)

(b)

B ×

× AB1 (1)

(c) B1

, isoscelesM1, A1

52 + 52 = (50)2, right-angled (or gradient method) M1, A1 (5)

(d) M1

= (–)AOB = M1, A1 (3)

(10 marks)[P4 June 2002 Qn 5]

7. ; at x = 2p M1, A1

Equation of tangent at P, M1

etc)

At Q q2y + x + 4q Two correct equations in any form A1

(p2 – q2)y = 4(p – q) M1

() A1

() M1, A1 (8)

[*P5 June 2002 Qn 7]

8. (a) For n = 1 25 + 52 = 57, which is divisible by 3 M1, A1

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 4

Assume true for n = k (k + 1)th term is 23k +5 + 5k+2 B1

(k + 1)th term kth term = 23k +5 + 5k+2 23k +2 + 5k+1 M1

= 23k +2(23 1) + 5k+1(5 1) M1, A1

= 6(23k +2 + 5k+1) + 3. 23k +2 or = 4(23k +2 + 5k+1) + 3. 23k +2 M1

which is divisible by 3 (k + 1)th term is divisible by 3 A1

Thus by induction true for all n cso B1 (9)

(b) For n = 1 RHS = B1

Assume true for n = k

M1 A3/2/1/0

(-1 each error)

B1

If true for k then true for k + 1 by induction true for all n B1 (7)

(16 marks)

[P6 June 2002 Qn 6]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 5

3.

9. f(2) = 1.514

B1

f() = 1.142 B1

M1

1.514 + 2 1.142 = (1.142 + 1.514)

= 2.65 A1 (4)[*P4 January 2003 Qn 4]

10. (a) z2 = (3 – 3i)(3 – 3i) = 18i M1 A1 (2)

(b) = = = M1 A1 (2)

(c) z = (9 + 9) = 18 = 32

z = 18 two correct M1

= = = all three correct A1 (2)

(d) C D O

two correct A

four correct

B

B1

B1 (2)

(e) = 18 M1 A1

AOB = COD = 45 similar B1 (3)

(11 marks)[P4 January 2003 Qn 6]

11. (a) x3 – 27 = (x – 3)(x2 + 3x + 9) M1

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 6

2 1.142

1.514

(x = 3 is one root). Others satisfy (x2 + 3x + 9) = 0 (*) A1 (2)

(b) Roots are x = 3 B1

and x = M1

= A1 (3)

(c)

3

3 and one other root in

correct quad

Root in complex

conjugate posn.

B1

B1 ft (2)

(7 marks)[#P4 June 2003 Qn 3]

12. (a) f(1) = –4 f(2) = 1 M1

Change of sign (and continuity) implies (1, 2) A1 (2)

(b) f(1.5) = –2.3… B1

f(1.75) = –0.9…

f(1.875) = –0.03… B1 (2)

NB Exact answer is 1.8789…

Alt to (a) y y = 3x

98 y = x + 67

3

O 1 2 x

Two graphs with single point of intersection (x > 0)

Two calculations at both x = 1 and x = 2

M1

A1

[*P4 June 2003 Qn 4]

13. (a) = = ( = 1 – i ) M1

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 7

| z | = , = awrt 1.12, accept exact equivalentsM1, A1

(3)

(b) = accept in (a) if clearly applied to (b)

M1

1 = obtaining quadratic or equivalent M1 A1

a2 + 5a – 6 = (a + 6)(a – 1)

a = –6, 1 M1 A1

Reject a = –6, wrong quadrant/– , one value A1 (6)

(9 marks)

Alt. (a) | 4 + 3i | = 5, | 2 + 4i | = 20 M1

| z | = ( = )M1 A1

(3)

(b) arg z = arg (a + 3i) – arg (2 + ai)

M1

1 = M1 A1

(3)

leading to a2 + 5a – 6 = 0, then as before

[P4 June 2003 Qn 5]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 8

[P6 June 2003 Qn 2]

15. (a) | z | = 22 | w | = 2 M1, A1

| wz2 | = (22)2 × 2 = 16 M1, A1

arg z = – arg w = ; arg wz2 = – – + = , 60° M1, A1 (6)

ALT z2 = –8i; z2w = 8 + 83i M1, A1

| z2w | = M1

= 16 A1

arg z2w = tan–13 M1

= A1

(b) C

B

A

Points A and B B1

Point C B1ft

angle BOC = – M1

= , 90° A1 (4)

(10 marks)[P4 January 2004 Qn 3]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 9

14.

[P4 June 2004 Qn 1]

17. and B1

B1

(2)

[*P4 June 2004 Qn 2]

[P4 June 2004 Qn 3]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 10

16.

18.

[P4 June 2004 Qn 5]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 11

19.

20.

The normal to the curve has gradient .

The equation of the normal is

The equation may be written

M1 A1

B1

M1A1 (5)

[*P5 June 2004 Qn 8]

21. (a) , M1

M1 A1 (3)

(b) M1

M1

In correct quadrant A1 (3)

(c) A B1 B B1 C ft B1ft (3)

(d) B1 Correct method for M1 A1 (3)

(e) awrt 3.46 M1 A1 (2)

[#FP1/P4 January 2005 Qn 8]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 12

A

B

C

22. (a) Long division or any complete method M1

M1 A1 (3)

(b)

In correct quadrants and on negative B1ft (1) - axis only required

(c) Full method M1

triangle is right angled A1 (2) (6)

Alternative to (c) Result follows by (converse of) Pythagoras, or any complete method. M1 A1

[#FP1/P4 June 2005 Qn 2]

23. … to 1sf or better B1

… Attempt at , M1A1 (3)

[*FP1/P4 June 2005 Qn 4]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 13

[FP1/P4 June 2005 Qn 5]

[*FP2/P5 June 2005 Qn 5]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 14

24.

25.

[FP3/P6 June 2005 Qn 1]

[FP1/P4 January 2006 Qn 1]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 15

26.

27.

[FP1/P4 January 2006 Qn 3]

29. (a) … awrt B1

… awrt B1

or equivalent M1

cao A1 (4)

(b) 112 (min) (1 hr 52 min) B1 (1) (5)

[#*FP1/P4 January 2006 Qn 5]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 16

28.

[*FP2/P5 Jan 2006 Qn 9]

[*FP3/P6 Jan 2006 Qn 3]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 17

30.

31.

33 . (a)

Adding Eliminating either variable M1

A1

M1

= A1 (4)

(b) arctan 2 M1

cao A1 (2)

(6 marks)[FP1 June 2006 Qn 1]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 18

32.

[*FP3/P6 January 2006 Qn 5]

34. (a) accept 1sf M1

Change of sign (and continuity) A1 (2)

(b) accept 1sf M1

M1 A1 (3)

(5)[*FP1 June 2006 Qn 6]

[FP1 Jan 2007 Qn 1]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 19

35.

[FP1 Jan 2007 Qn 3]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 20

36.

[FP1 June 2007 Qn 4]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 21

37.

[FP1 June 2007 Qn 6]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 22

38.

39. (a) Can be implied B1

Use of any correct method or formula to obtain an equation of PQ in any form. M1

Leading to A1 (3)

(b) Gradient of normal at P is . Given or implied at any stage B1 Obtaining any correct form for normal at either point. Allow if just written down. M1 A1

Using both normal equations and eliminating x or y. Allow in any unsimplified form. M1

Any correct form for x or y A1 Leading to cso A1 cso A1 (7)

[*FP2 June 2007 Qn8]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 23

40. : B1

(Hence result is true for .)

, by induction

hypothesisM1

M1 A1

(Hence, if result is true for , then it is true for )

By Mathematical Induction, above implies the result is true for all

cso A1 (5)

(5 marks) [FP3 June 2007 Qn5]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 24

41. (a)

M1

can be implied A1

M1

Hence cso A1 (4)

Note: is divisible by 240 and other appropriate multiples of 15 lead to the required result.

(b) : B1

(Hence result is true for .)

From (a) say. By induction hypothesis say.

M1

(Hence, if result is true for , then it is true for )

By Mathematical Induction, above implies the result is true for all

Accept equivalent arguments cso A1 (3)

(7 marks)[*FP3 June 2007 Qn6]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 25

[FP1 January 2008 Qn 2]

43. 3 dp or better B1, B1

Using to obtain M1

3dp or better A1 (4)

[*FP1 January 2008 Qn 4]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 26

42.

[FP1 January 2008 Qn 6]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 27

44.

[FP1 June 2008 Qn 1]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 28

45.

[FP1 June 2008 Qn 3]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 29

46.

[FP3 June 2008 Qn 5]

FP1 question mark schemes from old P4, P5, P6 and FP1, FP2, FP3 papers – Version 2 – March 2009 30

47.